Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $z = \dfrac{n^3 - 14n^2 + 45n}{-3n^2 + 21n - 30} \div \dfrac{n^2 - 6n}{-2n + 4} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $z = \dfrac{n^3 - 14n^2 + 45n}{-3n^2 + 21n - 30} \times \dfrac{-2n + 4}{n^2 - 6n} $ First factor out any common factors. $z = \dfrac{n(n^2 - 14n + 45)}{-3(n^2 - 7n + 10)} \times \dfrac{-2(n - 2)}{n(n - 6)} $ Then factor the quadratic expressions. $z = \dfrac {n(n - 5)(n - 9)} {-3(n - 5)(n - 2)} \times \dfrac {-2(n - 2)} {n(n - 6)} $ Then multiply the two numerators and multiply the two denominators. $z = \dfrac { n(n - 5)(n - 9) \times -2(n - 2)} { -3(n - 5)(n - 2) \times n(n - 6)} $ $z = \dfrac {-2n(n - 5)(n - 9)(n - 2)} {-3n(n - 5)(n - 2)(n - 6)} $ Notice that $(n - 5)$ and $(n - 2)$ appear in both the numerator and denominator so we can cancel them. $z = \dfrac {-2n\cancel{(n - 5)}(n - 9)(n - 2)} {-3n\cancel{(n - 5)}(n - 2)(n - 6)} $ We are dividing by $n - 5$ , so $n - 5 \neq 0$ Therefore, $n \neq 5$ $z = \dfrac {-2n\cancel{(n - 5)}(n - 9)\cancel{(n - 2)}} {-3n\cancel{(n - 5)}\cancel{(n - 2)}(n - 6)} $ We are dividing by $n - 2$ , so $n - 2 \neq 0$ Therefore, $n \neq 2$ $z = \dfrac {-2n(n - 9)} {-3n(n - 6)} $ $ z = \dfrac{2(n - 9)}{3(n - 6)}; n \neq 5; n \neq 2 $